contest-matheuclidean-geometrygeometrysolution-verificationtrigonometry
The problem is as the title suggests, in the figure given below, solve for the missing angle labeled "?". This problem is from my friend @geometri_hayattir on Instagram.
I have solved the problem, and I will share my approach as an answer below. Since the solution of this problem was not posted, I'm unsure if my approach and answer are correct. Please let me know if there are any issues in my method, and please share your own answers and approaches!
Your solution looks correct to me.
I'm going to write two solutions.
Solution 1 :
Let us take $E$ such that $\triangle{ACE}$ is an equilateral triangle. ($E$ is on the same side of $AC$ as $D$.)
Now, let us first show that $B,D,E$ are collinear.
Since $AB=AC=AE$, we see that $\triangle{ABE}$ is an isosceles triangle, and so we get $$\angle{ABE}=\frac{1}{2}(180^\circ-\angle{BAE})=\frac{1}{2}(180^\circ-100^\circ)=40^\circ=\angle{ABD}$$ Therefore, it follows that $B,D,E$ are collinear.
Since $\triangle{DCA}$ is congruent to $\triangle{DCE}$, we obtain $$x=\angle{DAC}=\angle{DEC}=\color{red}{20^\circ}$$
Solution 2 :
This is a trigonometric approach.
Using the law of sines in $\triangle{ABD}$, we have $$AD=\frac{\sin 40^\circ}{\sin(100^\circ-x)}AB\tag1$$ Using the law of sines in $\triangle{ADC}$, we have $$AD=\frac{\sin 30^\circ}{\sin(150^\circ-x)}AC\tag2$$
From $(1)(2)$ with $AB=AC$, we get $$\frac{\sin 40^\circ}{\sin(100^\circ-x)}=\frac{\sin 30^\circ}{\sin(150^\circ-x)}$$ i.e. $$\sin 40^\circ\ \bigg(\frac 12\cos x+\frac{\sqrt 3}{2}\sin x\bigg)=\frac 12(\sin 100^\circ\cos x-\cos 100^\circ\sin x)$$
Dividing the both sides by $\frac 12\cos x$, and solving it for $\tan x$ give $$\begin{align}\tan x&=\frac{\sin 100^\circ-\sin 40^\circ}{\sqrt 3\sin 40^\circ+\cos 100^\circ} \\\\&=\frac{\sin(90^\circ +10^\circ)-\sin(30^\circ+10^\circ)}{\sqrt 3\sin(30^\circ+10^\circ)+\cos(90^\circ +10^\circ)} \\\\&=\frac{\frac 12\cos 10^\circ-\frac{\sqrt 3}{2}\sin 10^\circ}{\frac{\sqrt 3}{2}\cos 10^\circ+\frac 12\sin 10^\circ} \\\\&=\frac{\frac{1}{\sqrt 3}-\tan 10^\circ}{1+\frac{1}{\sqrt 3}\tan 10^\circ} \\\\&=\frac{\tan 30^\circ -\tan 10^\circ}{1+\tan 30^\circ \tan 10^\circ} \\\\&=\tan (30^\circ -10^\circ) \\\\&=\tan 20^\circ\end{align}$$ Therefore, we get $x=\color{red}{20^\circ}$.
This is my approach to this problem:
Extend $BC$ and $AD$ such that they intersect at point $E$. Since $\angle ECD=90$ and $\angle EDC=30$, we can conclude that $\angle AEB=60$, $\triangle AEB$ is equilateral and $\triangle CED$ is a $30-60-90$ triangle. Let $DE=2a$, this means that $CE=a$. Since $\triangle AEB$ is equilateral we can say that:
$$48+2a=60+a$$
$$a=12$$
Therefore, $x=60+a=60+12=72$
Best Answer
This is my approach:
1.) Locate point $E$ outside of $\triangle ABC$ and join it via $AE$ and $CE$ such that $\angle EAC=\angle DAC=30^\circ$ and $AE=AD$. Notice that this means $\triangle AEC$ and $\triangle ADC$ are congruent via the SAS property. [See "motivation" at the end of the answer if you wish to know the motivation behind this particular construction].
2.) Notice that $\triangle AED$ is an equilateral triangle. Also notice that $\angle AEC=\angle ABC=130^\circ$. This implies that Quadrilateral $ACEB$ is a cyclic quadrilateral. Therefore, $\angle AEB=\angle ACB=10^\circ$ and $\angle BAE=\angle BCE=10^\circ$. This implies that $AB=BE$ and $\angle DAB=\angle DEB=70^\circ$. This proves that $\triangle DAB$ is congruent to $\triangle DEB$, therefore $\angle ADB=\angle EDB=30^\circ$. Therefore the missing angle is $60^\circ$.
[Motivation for (1): The motivation behind this construction is very simple, notice that $\angle ADC=\angle ABC=130^\circ$, so by "flipping" $\triangle ADC$ to a congruent triangle $\triangle AEC$ means that we can obtain a cyclic quadrilateral, which from where we can easily exploit the properties of the cyclic quadrilateral to obtain some angles]